Solve the equation $\frac { 20 } { 1 + 2 ^ { x } } = x$ Restrict your attention to the horizontal span of 0 to 5.
A) 4.15
B) 3.14
C) 2.69
D) 4.44

Question

Quizplus · Accepted Answer

We start by noticing that the left side of the equation is a decreasing function and the right side is an increasing function. Thus, there can be at most one solution.

Next, we observe that $x=0$ is not a solution, and $x=1$ gives $10=1$, which is not true. Therefore, the solution is in the interval $(1,5)$.

We can estimate the solution by noticing that for large $x$, the left side is very close to $\frac{20}{2^x}$, which equals $5\cdot 2^{-x}$. Thus, we want to solve $5\cdot 2^{-x}\approx x$, which implies that $x$ should be a bit larger than $2.5$ (since for $x=2.5$ we have $5\cdot 2^{-x}=2.5$), and therefore we expect the solution to be close to choice (C).

To get a more accurate solution, we can use a numerical method such as Newton-Raphson. We define $f(x)=\frac{20}{1+2^x}-x$, then $f'(x)=\frac{-20\ln2\cdot 2^x}{(1+2^x)^2}-1$. Starting with $x_0=2.5$, the Newton-Raphson iteration gives $x_1=2.694$, $x_2=2.692$, and so on. Therefore, the solution is $x\approx\boxed{	extbf{(C) }2.69}$.

Solve the Equation 201+2x=x\frac { 20 } { 1 + 2 ^ { x } } = x1+2x20​=x

Solve the Equation $\frac { 20 } { 1 + 2 ^ { x } } = x$